# 584. 寻找用户推荐人
# https://leetcode-cn.com/problems/find-customer-referee/

# SQL架构
Create table If Not Exists Customer (id int, name varchar(25), referee_id int);
Truncate table Customer;
insert into Customer (id, name, referee_id) values ('1', 'Will', null);
insert into Customer (id, name, referee_id) values ('2', 'Jane', null);
insert into Customer (id, name, referee_id) values ('3', 'Alex', '2');
insert into Customer (id, name, referee_id) values ('4', 'Bill', null);
insert into Customer (id, name, referee_id) values ('5', 'Zack', '1');
insert into Customer (id, name, referee_id) values ('6', 'Mark', '2');

# Write your MySQL query statement below
select
    name
from
    Customer
where
    referee_id != 2
    or referee_id is null;

# clean-up
drop table Customer;